Lecture 13

Lecture 13 - Rate Laws •  Rate laws are always determined experimentally unless you have been told a rate-limiting or slow step •  Reaction

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Unformatted text preview: Rate Laws •  Rate laws are always determined experimentally unless you have been told a rate-limiting or slow step. •  Reaction order is always defined in terms of reactant (not product) concentrations. •  The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation. ln[A]t =  ­kt + ln[A]0 1/[A]t = kt + 1/[A]0 [A]t =  ­kt + [A]0 Figure 16.5 Integrated rate laws and reac=on orders. ln[A]t =  ­kt + ln[A]0 1/[A]t = kt + 1/[A]0 [A]t =  ­kt + [A]0 Figure 16.6 Graphical determina=on of the reac=on order for the decomposi=on of N2O5. Figure 16.7 A plot of [N2O5] vs. =me for three half ­lives. for a first ­order process t1/2 = ln 2 k = 0.693 k •  Half ­life is defined as the =me required for one ­half of a reactant to react. •  Because [A] at t1/2 is one ­half of the original [A], [A]t = 0.5 [A]0. Sample Problem 16.6 PROBLEM: Using Molecular Scenes to Determine Half ­Life Compound A (red) converts to compound B (black) in a first ­order process. The following scenes represent the reac=on mixture at the start of the reac=on (at 0.0 s) and aaer 30.0 s: (a)  Find the half ­life, t1/2, of the reac=on. (b)  Calculate the rate constant, k. (c)  Draw a scene that represents the reac=on mixture aaer 2.00 min. PLAN: (a)  Half ­life is constant for first ­order reac=on. The =me it takes for half of the red spheres to become black is the half ­life. (b)  We subs=tute the value of t1/2 from part (a) into Equa=on 16.7 and solve for k. (c)  Knowing the half ­life, one can calculate how many red and black spheres would be in the mixture at 2.00 min. Sample Problem 16.6 Using Molecular Scenes to Determine Half ­Life con=nued SOLUTION: (a) (b) At t = 0.0 s, there are 8 A (red) molecules and no B (black). At t = 30.0 s, there are 6 A and 2 B molecules present, so ¼ of A has reacted. Therefore, t1/2 = (2 x 30.0 s) = 60.0 s For first ­order reac=on, t1/2 = 0.693/k k = 0.693/t1/2 = 0.693/60.0 s = 1.16 x 10 ­2 s ­1 (c) The =me given, t = 120. s, is two half ­lives. There should be ½(1/2) molecules of A (red) lea. Sample Problem 16.7 PROBLEM: Determining the Half ­Life of a First ­Order Reac=on Cyclopropane is the smallest cyclic hydrocarbon. Because its 60o bond angles reduce orbital overlap, its bonds are weak. As a result, it is thermally unstable and rearranges to propene at 1000oC via the following first ­order reac=on: The rate constant is 9.2 s ­1, (a) What is the half ­life of the reac=on? (b) How long does it take for the concentra=on of cyclopropane to reach one ­quarter of the ini=al value? 0.693 PLAN: Use the half ­life equa=on, t1/2 = , to find the half ­life. k One ­quarter of the ini=al value means two half ­lives have passed. SOLUTION: (a) 0.693 t1/2 =  ­ 1 = 0.075 s 9.2 s (b) 2 t1/2 = 2(0.075 s) = 0.15 s Half ­Life ­ 2nd order For a second ­order process, set [A]t=0.5 [A]0 in 2nd order equa=on. Figure 16.8 Dependence of the rate constant on temperature. The Effect of Temperature on Reac=on Rate The Arrhenius Equa=on where k is the kine=c rate constant at T Ea is the ac=va=on energy R is the universal gas constant ln k = ln A  ­ Ea/RT ln k2 k1 =  ­ Ea R T is the Kelvin temperature A is the collision frequency factor 1  ­ T2 1 T1 Figure 16.9 Graphical determina=on of the ac=va=on energy. ln k =  ­Ea/R (1/T) + ln A Sample Problem 16.8 PROBLEM: Determining the Energy of Ac=va=on The decomposi=on of hydrogen iodide, 2HI(g) H2(g) + I2(g) has rate constants of 9.51 x 10 ­9 L/mol•s at 500. K and 1.10 x 10 ­5 L/mol•s at 600. K. Find Ea. PLAN: Use the modifica=on of the Arrhenius equa=on to find Ea. SOLUTION: ln k2 k1 = – Ea = –(8.314 J/mol K) Ea 1 R T2 ln – 1 T1 1.10x10 ­5 L/mol s 9.51x10 ­9 L/mol s Ea = 1.76 x 105 J/mol = 176 kJ/mol k2 Ea = – R (ln ) k1 1 600 K – 1 – T2 1 500 K 1 T1  ­1  ­1 Figure 16.10 Informa=on sequence to determine the kine=c parameters of a reac=on. The dependence of number of possible collisions on the product of reactant concentra=ons. The Collision Theory •  In a chemical reac=on, bonds are broken and new bonds are formed. •  Molecules can only react if they collide with each other. Figure 16.12 The effect of temperature on the distribu=on of collision energies. Figure 16.13 Energy ­level diagram for a reac=on. The forward reac=on is exothermic because the reactants have more energy than the products. Figure 16.14 The importance of molecular orienta=on to an effec=ve collision. NO + NO3 2 NO2 A is the frequency factor A = pZ where Z is the collision frequency p is the orienta=on probability factor Figure 16.15 Nature of the transi=on state in the reac=on between CH3Br and OH ­. CH3Br + OH ­ CH3OH + Br  ­ transi7on state or ac7vated complex Figure 16.16 Reac=on energy diagram for the reac=on of CH3Br and OH ­. Figure 16.17 Reac=on energy diagrams and possible transi=on states for two reac=ons. ...
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This note was uploaded on 02/18/2012 for the course CHEM 6C taught by Professor Hoeger during the Spring '08 term at UCSD.

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